Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long int ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T bpow(T p, T e) {
  unsigned long long int ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p);
    p = (p * p);
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
int toInt(string s) {
  int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
int toLlint(string s) {
  long long int sm;
  stringstream ss(s);
  ss >> sm;
  return sm;
}
long long int ts, kk = 1;
int n, m;
int gr[105][105];
int cl[105];
int rw[105];
int nw[105];
int rq[105];
int clc(int c, int x) {
  int i, j, k;
  for (i = 0; i < n; i++) rq[i] = gr[i][c] - x;
  for (j = 0; j < m; j++) nw[j] = 505;
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (j == c) continue;
      nw[j] = min(nw[j], gr[i][j] - rq[i]);
      if (nw[j] < 0) return 1000000007LL;
    }
  }
  for (i = 0; i < n; i++) {
    for (j = 0; j < m; j++) {
      if (j == c) continue;
      if (gr[i][j] != nw[j] + rq[i]) return 1000000007LL;
    }
  }
  int rs = 0;
  nw[c] = x;
  for (i = 0; i < n; i++) rs += rq[i];
  for (j = 0; j < m; j++) rs += nw[j];
  return rs;
}
int main() {
  int t, i, j, k, l;
  scanf("%d%d", &n, &m);
  for (i = 0; i < m; i++) cl[i] = 505;
  t = 0;
  for (i = 0; i < n; i++) {
    rw[i] = 505;
    for (j = 0; j < m; j++) {
      scanf("%d", &k);
      t = max(t, k);
      gr[i][j] = k;
      rw[i] = min(rw[i], k);
      cl[j] = min(cl[j], k);
    }
  }
  int q1, q2;
  q1 = q2 = 0;
  for (j = 0; j < n; j++) q1 += rw[j];
  for (j = 0; j < m; j++) q2 += cl[j];
  if (q1 + q2 == 0) {
    if (t != 0)
      t = -1;
    else
      t = 0;
    printf("%d\n", t);
    return 0;
  }
  if (q1 == 0) {
    for (i = 0; i < n; i++)
      for (j = 0; j < m; j++) {
        if (gr[i][j] != cl[j]) {
          printf("-1\n");
          return 0;
        }
      }
    printf("%d\n", q2);
    for (j = 0; j < m; j++) {
      for (i = 0; i < cl[j]; i++) printf("col %d\n", j + 1);
    }
    return 0;
  }
  if (q2 == 0) {
    for (i = 0; i < n; i++)
      for (j = 0; j < m; j++) {
        if (gr[i][j] != rw[i]) {
          printf("-1\n");
          return 0;
        }
      }
    printf("%d\n", q1);
    for (j = 0; j < n; j++) {
      for (i = 0; i < rw[j]; i++) printf("row %d\n", j + 1);
    }
    return 0;
  }
  for (j = 0; j < m; j++)
    if (cl[j]) break;
  int rs = 1000000007LL;
  int id;
  for (i = 0; i <= cl[j]; i++) {
    int q = clc(j, i);
    if (q < rs) {
      rs = q;
      id = i;
    }
  }
  if (rs == 1000000007LL) {
    printf("-1\n");
    return 0;
  }
  printf("%d\n", clc(j, id));
  for (i = 0; i < n; i++)
    for (k = 0; k < rq[i]; k++) printf("row %d\n", i + 1);
  for (i = 0; i < m; i++)
    for (k = 0; k < nw[i]; k++) printf("col %d\n", i + 1);
  return 0;
}